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Given RMS Speed, Find Temperature

  • Thread starter asifion
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Problem Statement
You are watching a science fiction movie in which the hero shrinks down to the size of an atom and fights villains while jumping from air molecule to air molecule. In one scene, the hero's molecule is about to crash head-on into the molecule on which a villain is riding. The villain's molecule is initially 50 molecular radii away and, in the movie, it takes 3.0 s for the molecules to collide. Estimate the air temperature required for this to be possible. Assume the molecules are nitrogen molecules, each traveling at the rms speed.
Relevant Equations
rms speed=√((3k_B * T)/m)
mass of N2 = 28 * 1.67e-27 kg
not sure if this is the right one - just googled it
triple covalent bond N2 radii = 54e-12 m

I tried to first find the rms speed:

v = x/t
= 50 * 54e-12 / 3 m/s

Then I solved for T (in K):
242480

[50 * 54e-12 / 3]^2 * [28 * 1.67e-27] / [3 * 1.38e-23] = 9.14e-22 K
 
Last edited:

TSny

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Both molecules are in motion heading directly toward each other with the same speed. How far will each molecule move before the collision occurs? So, what should you use for x in your calculation of v? (You didn't include the units for the bond radius or for the temperature in your answer.)

I guess you have to assume that there has been no scaling of time in the movie. No information is given in that regard. So, I think you are OK with using t = 3.0 s in determining v.

Does it surprise you that you are getting an extremely cold temperature? At room temperature, what is the approximate value of vrms for nitrogen? At room temperature, approximately how long would it take for the two molecules in the movie to collide?
 
I added the units to the original question for future reference.

Ok, so since they are headed towards each other at the same speed, they should meet at the middle thus only travelling half the distance.

I tried [(50/2) * 54e-12 / 3]^2 * [28 * 1.67e-27] / [3 * 1.38e-23] and got 2.28e-22 K but that still isn't correct.
 
No, it should be a cold temperature since it takes them 3 whole seconds to travel such a such distance.
 

TSny

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I added the units to the original question for future reference.

Ok, so since they are headed towards each other at the same speed, they should meet at the middle thus only travelling half the distance.

I tried [(50/2) * 54e-12 / 3]^2 * [28 * 1.67e-27] / [3 * 1.38e-23] and got 2.28e-22 K but that still isn't correct.
Are you way off? Do you know the value that you should get?
 
Are you way off? Do you know the value that you should get?
No, I don't know the value I should get. But in the next part, the correct answer is that this is not a plausible temperature for air.
 

TSny

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What happens to air (nitrogen) at a temperature of about 75 K at normal pressure?
 
What happens to air (nitrogen) at a temperature of about 75 K at normal pressure?
In terms of what? I'd believe that since it is just above the freezing point of air, the molecules will move slower and be more compacted.
 

TSny

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At normal pressure, nitrogen becomes a liquid at about 77 K and a solid at around 63 K. But in the scenario of the movie, the nitrogen is a gas.
 
Shouldn't the velocity be ##x/2*t## because it is you are measuring time and distance? But relative to you the molecule is moving with ##2v## velocity.
 

TSny

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Shouldn't the velocity be ##x/2*t## because it is you are measuring time and distance? But relative to you the molecule is moving with ##2v## velocity.
I believe this was addressed in posts #2 and #3.
 
At normal pressure, nitrogen becomes a liquid at about 77 K and a solid at around 63 K. But in the scenario of the movie, the nitrogen is a gas.
How does that affect the calculations and answer?
 

TSny

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Nitrogen cannot exist as a gas below about 75 K. So, your answer for the temperature that you got in post #3 would support the statement that your answer is "not a plausible temperature for air".
 
Nitrogen cannot exist as a gas below about 75 K. So, your answer for the temperature that you got in post #3 would support the statement that your answer is "not a plausible temperature for air".
Yes, I understand that, but do you know why the calculation is incorrect?
 

TSny

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Yes, I understand that, but do you know why the calculation is incorrect?
No, I don't see why it's incorrect. But I also don't understand why you say it's incorrect :oldsmile:
 
The answer says incorrect.
 

TSny

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Bear with me, please. Are you entering your numerical answer for the temperature into some sort of homework software and getting the response of "incorrect"?
 
Yes, I am; sorry for not being more specific.
 

TSny

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OK. Thanks. So, we don't have any idea if your answer is in the ball-park. You used the "bond radius" for the radius of the molecule. This seems to me to be somewhat of an underestimate. A better estimate for the size of the molecule might be something called the "kinetic" size. See here:

There is a table that lists the "kinetic diameter" for nitrogen. If you use half of this for the radius, then you will find that it will change the answer for the temperature by roughly a factor of 10. But I don't know if this is going to make the difference.

In post #6 you mentioned another part of the problem for which the answer is something like "this is not a plausible temperature for air." Would you mind quoting the problem statement for this part? It might help.
 
I tried the kinetic radius, which also didn't work.

Is this a plausible temperature for air? Yes or No
No was the correct answer.
 

TSny

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I tried the kinetic radius, which also didn't work.
OK. Sorry, I don't see where to go from here.

Is this a plausible temperature for air? Yes or No
No was the correct answer.
This would, of course, be true for your answer for the temperature. But it doesn't help with getting the right numerical answer for the temperature.

Maybe someone else will chime in with some ideas.
 

TSny

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@asifion Since T depends on the square of the speed, the answer for T is pretty sensitive to the choice of the molecular radius. The problem asks for an estimate of T. I wonder what tolerance is allowed in the answer.

Anyway, if you discover how to get the correct answer, I would appreciate it if you would post it here; especially if the answer is very different from what you got above. Thanks.
 

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